Formula of the Day: CCNpqCCNpNqp.

Formula of the day: CCCpqrCsCCqCrtCqt

The last two formulas of the day were CpCqp and CCpCqrCCpqCpr.  Together these form a set of axioms sufficient for the positive implicational calculus (equivalently they form a "basis" for the positive implicational calculus).  C. A. Meredith figured out that CCCpqrCsCCqCrtCqt itself serves as a basis for the positive implicational calculus of propositions.  To prove that{CpCqp-RVP, CCpCqrCCpqCpr-SD} yields CCCpqrCsCCqCrtCqt we can construct a demonstration as follows, with Ci standing for conditional introduction:

hypothesis  1 ! CCpqr
hypothesis  2 !@ s
hypothesis  3 !@# CqCrt
hypothesis  4 !@#$ q
RVP            5 !@#$ CqCpq
D5.4            6 !@#$ Cpq
D1.6            7 !@#$ r
D3.4            8 !@#$ Crt
D8.7            9 !@#$ t
Ci 4-9        10 !@# Cqt
Ci 3-10      11 !@  CCqCrtCqt
Ci 2-11      12 !     CsCCqCrtCqt
Ci 1-12      13  CCCpqrCsCCqCrtCqt.

We can expand all of this out and write a formal proof of {CCCpqrCsCCqCrtCqt} from the basis {CpCqp-RVP, CCpCqrCCpqCpr-SD} using any decent proof of the deduction metatheorem.  Here the letters a, b, ..., n, o can get substituted by something else, but p, q, ..., y, z cannot.  First we'll take out the "$" lines.

hypothesis   1 ! CCpqr
hypothesis   2 !@ s
hypothesis   3 !@# CqCrt
D[RVP].1    4 !@# CaCCpqr This is C4-1
D[SD].4      5 !@# CCaCpqCar
D5.RVP       6 !@# Cqr  ... C4-7
D[RVP].3    7 !@# CaCqCrt ... C4-3
D[SD].7      8 !@# CCaqCaCrt
D8.[Caa]     9 !@#  CqCrt ... C4-8
D[SD].9    10 !@# CCqrCqt
D10.6        11 !@# Cqt  C4-9
.
.
.

Now for the "#" lines in the above demonstration.  C3-3 and C3-9 match exactly.  C3-10 we can obtain simply from SD just by substitution.

hypothesis   1 ! CCpqr
hypothesis   2 !@ s
theorem       3 !@ CCaCbCcdCaCCbcCbd (we'll use this when we used "SD" above)
D[RVP].1    4 !@ CaCCpqr
D[RVP].4    5 !@ CaCbCCpqr ... C3-4
D3.5            6 !@ CaCCbCpqCbr ... C3-5
theorem       7 !@ CaCbCcb
D[SD].6      8 !@ CCaCbCpqCaCbr
D8.7            9 !@ CaCqr ... C3-6
D3.RVP    10 !@ CCabCCcaCcb  ... C3-8
D[SD].SD 11 !@ CCCaCbcCabCCaCbcCac
D11.9        12 !@ CCqCraCqa  C3-11

Now the "@" lines.

hypothesis    1 ! CCpqr
theorem        2 ! CCaCbCcdCaCCbcCbd
theorem        3 ! CCaCbCcCdeCaCbCCcdCce (we'll use this where we used 3 above)
D[RVP].1     4 ! CaCCpqr   C2-1
D[RVP].4     5 ! CaCbCCpqr  C2-4
D[RVP].5     6 ! CaCbCcCCpqr ... C2-5
D3.6             7 ! CaCbCCcCpqCcr ... C2-6
theorem        8 ! CaCbCcCdc ... C2-7
D2.7             9 ! CaCCbCcCpqCbCcr ... C2-8
D[SD].9     10 ! CCaCbCcCpqCaCbCcr
D10.8         11 ! CaCbCqr ... C2-9
theorem      12 ! CaCbCcb
D3.12         13 ! CaCCbcCCdbCdc  ... C2-10
D[SD].SD  14 ! CCCaCbcCabCCaCbcCac
D[RVP].14 15 ! CaCCCbCcdCbcCCbCcdCbd ... C2-11
D[SD].15   16 ! CCaCCbCcdCbcCaCCbCcdCbd
D16.11       17 ! CaCCqCrbCqr ... C2-12

And now we'll write the formal proof (allowing substitution for all lower case letters):

axiom     1 CpCqp   C1-4
axiom     2 CCpCqrCCpqCpr
D1.2       3 CpCCqCrsCCqrCqs
D2.3       4 CCpCqCrsCpCCqrCqs ... SD above
D1.4       5 CpCCqCrCstCqCCrsCrt
D2.5       6 CCpCqCrCstCpCqCCrsCrt   2 above
D1.6       7 CpCCqCrCsCtuCqCrCCstCsu
D2.7       8 CCpCqCrCsCtuCpCqCrCCstCsu ... 3 above
D1.1       9 CpCqCrq
D2.9     10 CCpqCpCrq
D10.1   11 CpCqCrp ... C1-5
D10.11 12 CpCqCrCsp ... C1-6
D1.9     13 CpCqCrCsr  ... C1-12
D1.13   14 CpCqCrCsCts ... C1-8
D8.12   15 CCpqCrCsCCtpCtq ... C1-7
D6.15   16 CCpqCrCCsCtpCsCtq ... C1-9
D4.16   17 CCpqCCrCsCtpCrCsCtq ... C1-10
D2.17   18 CCCpqCrCsCtpCCpqCrCsCtq
D18.14 19 CCCpqrCsCtCqr ... C1-11
D8.13   20 CpCqCCrsCCtrCts ... C1-13
D4.3     21 CpCCCqCrsCqrCCqCrsCqs ... C1-14
D1.21   22 CpCqCCCrCstCrsCCrCstCrt ... C1-15
D4.22   23 CpCCqCCrCstCrsCqCCrCstCrt ... C1-16
D2.23   24 CCpCqCCrCstCrsCpCqCCrCstCrt
D24.19 25 CCCpqrCsCCqCrtCqt

I'll point out that CCpqCCrpCrq is a theorem of this system as follows:

axiom   1 CpCqp
axiom   2 CCpCqrCCpqCpr
D1.2     3 CpCCqCrsCCqrCqs
D2.3     4 CCpCqCrsCpCCqrCqs
D4.1     5 CCpqCCrpCrq

But, even though it did appear in the set-ups to prove CCCpqrCsCCqCrtCqt, we never produced an actual proof of CCpqCCrpCrq, though we did produce a proof of  CpCqCCrsCCtrCts in which CCpqCrpCrq can get said to qualify as contained within.  I've used Prover9 to find a proof of {CpCqp, CCpCqrCCpqCpr} from {CCCpqrCsCCqCrtCqt}.  Meredith also found that CtCCpqCCCspCqrCpr can also serve as a basis of the positive implicational calculus.  Ted Ulrich has found some other single axioms for this system here.

Both axioms, "CtCCpqCCCspCqrCpr" and "CCCpqrCsCCqCrtCqt" have 17 symbols.  A stronger system, the implicational calculus of propositions, has a single axiom with just 13 symbols.

Also, the conversion procedure I used implicitly here depends on *how* a particular meta-proof of the deduction meta-theorem works.  If we have a different meta-proof of the deduction meta-theorem, we'll have a different conversion procedure.  My May 10th answer here shows the same process via a distinct meta-proof of the meta-deduction theorem.

Logical Formula of the Day: self-distribution CCpCqrCCpqCpr.

Here's two links on condensed detachment.  If we have two formulas Cab, and "a", condensed detachment produces the most general formula "b".  For instance, if we have

1. CpCqp as "a" and
2. CCpCqrCCpqCpr as "Cab",

we could substitute "r" in CCpCqrCCpqCpr, and thus obtain CCpqCpp from a condensed detachment D2.1.  We could also substitute "r" with "Cqp", and "p" with "Cqp" in  CCpCqrCCpqCpr  with Cqp  obtaining CCCqpCqCqpCCCqppCCqpCqp, and then substitute "p" with "Cqp" in CpCqp obtaining CCqpCqCqp.  Thus, from 1. and 2. we could detach CCCqpqCCqpCqp as a theorem.  However, CCCqpqCCqpCqp is less general than CCpqCpp, since if we substitute "p" with "Cqp" in CCpqCpp, then we obtain CCCqpqCCqpCqp.  So, condensed detachment Da.b obtains a form from which you can obtain every other theorem detachable from a substitution instance of "a" as the major premise and "b" as the minor premise.

Let's look a little at the system {CCpCqrCCpqCpr}

axiom   1 CCpCqrCCpqCpr
D1.1     2 CCCpCqrCpqCCpCqrCpr
D1.2     3 CCCCpCqrCpqCpCqrCCCpCqrCpqCpr
D1.3     4 CCCCCpCqrCpqCpCqrCCpCqrCpqCCCCpCqrCpqCpCqrCpr

Every theorem of this system obtained via condensed detachment has three variables.  This follows, because the axiom has only three variables in the antecedent and the consequent.  Thus, the second theorem has only three variables in the antecedent and the consequent.

If we have CCpCqrCCpqCpr and CpCqp as axioms or theorems, then we have a deduction metatheorem for the system which says "if we have an assumption "a" which leads to a formula "b" under the scope of some set of assumptions "g", then Cab holds under the scope of the assumptions "g" also".  Or more concisely we can write:

Deduction Metatheorem: If {g, a} |- b, then {g}|-Cab.

Formula of the Day: Recursive Variable Prefixing CpCqp.

When proving that a logical system has a deduction metatheorem, CpCqp proves indispensable.   In fact, CpCqp along with CCpCqrCCpqCpr or CCpqCCpCqrCpr suffices to prove a deduction metatheorem.  Let's consider the system {CpCqp}.  For this system:

Lemma: Every theorem obtainable by condensed detachment for the system {CpCqp} has the form C0C1...CnCpCqp.

Demonstration: The first theorem obtainable by condensed detachment is CpCqCrq.  Now from {CpCqp, CpCqCrq} we can use either thesis as the minor premise, and either one as the major.  If CpCqp is the major, then we will detach Cab where "a" is a variable and "b" consists of the minor premise.  If CpCqCrq is the major premise, then we will obtain CpCqp.  More generally, if we have CpCqp as the major premise and use any theorem "b" as the minor premise, we will obtain Cab upon a condensed detachment.  If Cat, where "a" does not appear in "t" gets used as the major premise, then we will obtain "t" after a condensed detachment.  Thus, any time we use condensed detachment in this system we'll either obtain CpCqp or some theorem which has CpCqp as its last 5 letters.

The above lemma can get argued also via mathematical induction.

Corollary: Every thesis (theorem or axiom) obtainable via condensed detachment for the system {CpCqp} can serve as a sole axiom for a system with the same theses as {CpCqp}.

Demonstration: Since every axiom has form C0C1...CnCpCqp by the above lemma, we can obtain CpCqp as a theorem of the system.  Since CpCqp has C0C1...CnCpCqp as a theorem, it follows that the systems have the exact same theorems.

Consequently it follows that in addition to {CpCqp, CCpCqrCCpqCpr} sufficing to prove the deduction theorem *any* theorem obtainable from the system {CpCqp} solely by condensed detachment can take the place of CpCqp in {CpCqp, CCpCqrCCpqCpr} .

Formula of the Day: CδCpqδANpq.

"δ" in opposition to representing a propositional variable indicates a functorial variable of arity one.  In other words, it stands for any of the unary functions presumed.  Given the context of classical logic this means it stands for any of the functions belonging to {F, N, I, V} where F(0)=0, F(1)=0, N(0)=1, N(1)=0, I(0)=0, I(1)=1, V(0)=1, V(1).  Thus we can check that CδCpqδANpq constitutes a tautology by checking that each member of {CFCpqFANpq, CNCpqNANpq, CICpqIANpq, CVCpqVANpq} is a tautology in 2-valued logic.

In notation for substitution for δ, an apostrophe " ' " means that we substitute the argument that follows delta for the apostrophe.  For example, from 1 δp, 1 δ/C'q we substitute δ with C, then the argument that follows δ which is "p" in this case, and then "q" obtaining Cpq.  If we write 2 CsδCpCqr, 2 δ/C'' that means we substitute δ with C, then the argument of δ CpCqr, then the argument of δ CpCqr obtaining CsCCpCqrCpCqr.  If we write 3 CpCCpδqr, 3 δ/CC'r' then we write CC, the argument of δ q, r, and lastly the argument of δ "q" again in place of δq obtaining CpCCpCCqrqr.  δ/' just means we substitute δ with the argument of δ.
CδCpqδANpq can effectively take the place of a definition of the conditional "C" in a logical calculus.  Let's consider CδCpqδANpq in the following system:

axiom                     1 ApNp
axiom                     2 CδCpqδANpq
2 δ/CCδCpqδ'CδANpqδCpq    3 CCCδCpqδCpqCδANpqδCpqCCδCpqδANpqCδANpqδCpq
2 δ/Cδ'δCpq               4 CCδCpqδCpqCδANpqδCpq
D3.4                      5 CCδCpqδANpqCδANpqδCpq
D5.2                      6 CδANpqδCpq
6 δ/'                     7 CANpqCpq
D7.1                      8 CpNNp

Thus, double negation introduction CpNNp follows from these two axioms.  It also follows that in the presence of a definition corresponding to axiom 2, the law of the excluded middle ApNp allows us to derive the law of double negation introduction CpNNp.

Well-Formed Formula of the Day: ApNp

ApNp consists of a symbolic formulation of the law of the excluded middle "either a proposition is true or it is not true."  If we wanted to use at as an axiom, we might want to have as a deduction rule {ANpq, p}->q, but even then we'd probably want to commute ApNp and use ANpp instead.  Let's say we have ANpp as an axiom along with ANpANqp as an axiom.  We might proceed as follows.

axiom  1 ANpp
axiom  2 ANpANqp
D2.2    3 ANpANqANrq
D2.1    4 ANpANqq

ApNp can get proved from the axiom set {CpCqp, CCpCqrCCpqCpr, CCNpqCCNpNqp, CpApq, CpAqp}.  ApNp is unique among classical logic theorems which only have C, A, K, E, and N as connectives in them in that it has exactly 4 symbols.

Formula of the Day: EEEpqrEpEqr

EEEpqrEpEqr comes as a way of saying that logical equivalence associates.  Along with EEpqEqp can serve as one of the axioms of the equivalential calculus.  Equivalential calculus allows us to deduce all tautologies which only have equivalence symbols and variables in them.  The rule of inference for equivalential calculus comes as a modified form of detachment "from Eab, as well as "a", we may infer b" where "a" and "b" consist of meta-variables.  Or more compactly

{Eab, a} -> b

We can condensed such an E-detachment along with uniform substitution just like we can condense C-detachment with uniform substitution.  Taking EEEpqrEpEqr as an axiom, we can develop the system {EEEpqrEpEqr} as follows:

axiom 1 EEEpqrEpEqr
D1.1   2 EEpqErEpEqr
D1.2   3 EpEqErEpEqr
D2.1   4 EsEEEpqrEEpEqrs
D1.3   5 EpEqErEsEEpqErs
D1.4   6 EpEqEEErstEErEstEpq
D2.2   7 EpEEqrEEsEqErsp
D2.3   8 EsEpEEqErEpEqrs

The system with a single axiom which corresponds to each of these theorems has no last theorem under condensed detachment (unlike {Cpp} which has Cpp as its only theorem under condensed detachment).  For {EEpqErEpEqr} each theorem will have some form Eab, which implies that there exists a substitution which enables a detachment from the antecedent of Epq of EEpqErEpEqr with EEpqErEpEqr as the major premise, and any theorem of the system {EEpqErEpEqr}.  For the systems corresponding to theorems 3 through 8, each axiom serving as the major premise and any theorem serving as the minor premise enables a condensed detachment.  Does *every* theorem EEEpqrEpEqr correspond to a system with a sole axiom which has no last theorem under condensed detachment?

EEEpqrEpEqr seems to come as a theorem that requires a fair amount of work to prove in the natural deduction systems I've encountered and in non-equivalential systems of logic which use definitions to get equivalences into the system.